I have posted the following question also here a longer time ago, but due to no answers I thought it might fit better to MO.

Let $(\Omega,\mathcal F)$ be a measurable space and $\mathcal P$ a weak*-compact subset of the set of all probability measures on $\mathcal F$. Does there exist a probability measure $\mathbb Q$ such that every measure $\mathbb P\in\mathcal P$ is absolutely continuous to $\mathbb Q$, i.e. such that $\mathbb P$ dominates the mesures in $\mathcal P$?

The weak*-topology is the weakest topology such that all the linear functionals $L_Z:\mathbb P\mapsto \int_\Omega Zd\mathbb P$ for $Z:\Omega\rightarrow\mathbb R$ $\mathcal F$-measurable and bounded, are continuous.

Note that when the set $\mathcal P$ is countable one can easily find a dominating measure for $\mathcal P$.

For future readers: as mentioned below, please note that the weak-* topology here is the one induced by considering measures on $(\Omega, \mathcal{F})$ as linear functionals on the space of bounded measurable functions on $\Omega$ (not continuous functions; $\Omega$ has not been given a topology.) For Mike Jury's comment, the space of all probability measures on $[0,1]$ is not compact in this topology, and for Gerald Edgar's, $x \mapsto \delta_x$ is not continuous.
–
Nate EldredgeMay 19 '13 at 13:28

4 Answers
4

First, given two finite measures $\mu,\nu$ on $(\Omega,\mathcal{F})$, the Lebesgue decomposition theorem says that there is an $A\in\mathcal{F}$ such that $1_{\Omega\setminus A}\mu$ is absolutely continuous with respect to $\nu$ and $1_A\mu,\nu$ are singular. This can also be constructed using the Radon-Nikodym derivative as $A=\lbrace d\nu/d(\mu+\nu)=0\rbrace$, and $A$ is uniquely defined up to a $\mu$-null set. We can think of $\mu(A)$ as measuring the distance $\mu$ is from being absolutely continuous wrt $\nu$, and I will write $d(\mu,\nu):=\mu(A)$.

Assuming $\mathcal{P}$ is nonempty then, using dependent choice, there exists a sequence $\mathbb{P}\_1,\mathbb{P}\_2,\ldots\in\mathcal{P}$ such that
$$
d(\mathbb{P}\_n,\mathbb{P}\_1+\cdots+\mathbb{P}\_{n-1}) \ge \frac12\sup\left\lbrace d(\mathbb{P},\mathbb{P}\_1+\cdots+\mathbb{P}\_{n-1})\colon\mathbb{P}\in\mathcal{P}\right\rbrace.
$$
We can show that $\mathbb{Q}:=\sum_{n=1}^\infty2^{-n}\mathbb{P}\_n$ is a dominating measure for $\mathcal{P}$.

Let me start by showing that $\alpha_n:=d(\mathbb{P}\_n,\mathbb{P}\_1+\cdots+\mathbb{P}\_{n-1})$ tends to zero. By definition, there exists $A_n\in\mathcal{F}$ such that $\mathbb{P}\_n(A_n)=\alpha_n$ and $\mathbb{P}\_m(A_n)=0$ for $m < n$. Replacing $A_n$ by $A_n\setminus\bigcup_{m > n}A_m$ if necessary, we can further suppose that $A_n$ are disjoint. Now, set $B_n=\bigcup_{m\ge n}A_m$ so that $\mathbb{P}\_m(B_n)\ge\alpha_m$ for $m\ge n$ and $B_n\downarrow\emptyset$ as $n\to\infty$. If $\alpha_n$ does not tend to zero then, by passing to a subsequence if necessary, it can be assumed that $\alpha_n\rightarrow\alpha > 0$. Now, by compactness, the sequence $\mathbb{P}\_m$ has a limit point $\mathbb{P}$ in the weak-* topology. By the definition of the topology given in the question, this means that $\mathbb{P}(B_n)$ is a limit point of $\mathbb{P}\_m(B_n)\ge\alpha_m$ as $m\to\infty$, so $\mathbb{P}(B_n)\ge\alpha$. This contradicts the fact that $\mathbb{P}(B_n)\to0$ as $n\to\infty$ (by countable additivity). So, $\alpha_n\to0$ as required.

Finally, lets show that $\mathbb{Q}$ is a dominating measure. Choosing any $\mathbb{P}\in\mathcal{P}$ set $\alpha:=d(\mathbb{P},\mathbb{Q})$. If $\alpha > 0$ then we would have $\alpha_n < \alpha/2$ for large $n$, in which case
$$
\begin{align}
d(\mathbb{P}\_n,\mathbb{P}\_1+\cdots+\mathbb{P}\_{n-1})&=\alpha_n < \alpha/2 =\frac12d(\mathbb{P},\mathbb{Q})\cr
&\le\frac12d(\mathbb{P},\mathbb{P}\_1+\cdots+\mathbb{P}\_{n-1}).
\end{align}
$$
The last inequality uses the fact that $\mathbb{P}\_1+\cdots+\mathbb{P}\_{n-1}$ is absolutely continuous wrt $\mathbb{Q}$. This inequality contradicts the choice of $\mathbb{P}\_n$, so we have $\alpha=0$ and $\mathbb{P}$ is absolutely continuous wrt $\mathbb{Q}$.

This is really beautiful!The only thing I have still to think about is the existence of such a sequence $(\mathbb P_n){n\in\mathbb N}$...
–
andy teichMay 16 '13 at 22:12

@andy: By definition of the supremum, for each n you can choose $d(P_n,P_1+\cdots+P_{n-1})$ as close as you like to the supremum, so can certainly choose it to be at least 1/2 of it.
–
George LowtherMay 16 '13 at 22:19

A bit more can be said than in George Lowther's answer. See Bogachev's Measure Theory volume 1, page 291 for a thorough discussion: there exists a dominating measure $Q$ with respect to which every measure $P \in \mathcal{P}$ is uniformly absolutely continuous, meaning that for all $\epsilon > 0$ there exists $\delta > 0$ such that $P(A) \le \epsilon$ for all $P \in \mathcal{P}$ whenever $A$ is measurable and $Q(A) \le \delta$. Moreover, the set of densities $\{dP/dQ : P \in \mathcal{P}\}$ is weakly compact in $L^1(Q)$.

As noted in the OP, if $\mathcal P$ is countable, for example $\mathcal P=(\mu_n,n\in\Bbb N)$, then take $\nu:=\sum_{n=1}^{+\infty}2^{-n}\mu_n$.

Edit:This was posted before having the wanted definition of weak star topology

When $\mathcal P$ is not countable, it's not necessarily true: take $\Omega:=[0,1]$, $\mathcal F$ its Borel $\sigma$-algebra and $\mathcal P:=\(\delta_x,x\in[0,1]\)$. Then $\mathcal P$ is weak-$*$ star compact (as $x\mapsto \delta_x$ from $[0,1]$ to $\mathcal P$ is a homeomorphism), but there is no measure $\nu$ such that $\delta_x\ll\nu$ for all $x\in[0,1]$. Otherwise, $\nu(x)>0$ for all $x$, which is incompatible with additivity of $\nu$ and uncountability of the unit interval.

Considering the weak star topology as given in the OP, the answer seems to be yes. First, let $\mu\sim\nu$ mean that $\mu\ll\nu$ and $\nu\ll\mu$. We identify two equivalent measures. Then we work with Zorn's lemma with order $\mu\leq\nu$ iff $\mu\ll\nu$. Weak star compacteness allows us to show that each totally ordered subset of $\mathcal P$ admits a maximal element.

Does this really answer the intended Question? With the "weak*-topology" as defined above, the map $x\maspto \delta_x$ is not continuous.
–
Lutz MattnerMay 13 '13 at 15:52

1

@Lutz... you are right. Because andy has the wrong definition for the weak* topology. The usual way to define it is to use only continuous $Z$. If we use all measurable $Z$, as andy does, then you get a much stronger topology. Moreover, andy didn't even say that $Z$ should be bounded. So, let's give him a chance to say whether he wants to correct the definition.
–
Gerald EdgarMay 13 '13 at 16:56

Andy does not have a definition at all because his functionals $L_Z$ make sense only for bounded $Z$. Anyway, weak*-compactness could be obtained from the Banach-Alaoglu theorem.
–
Jochen WengenrothMay 13 '13 at 18:45

@Gerald Edgar: "...the usual way to define it is to use continuous $Z$." Maybe you realized that we are working on measurable spaces and there is no notion of continuity...
–
andy teichMay 14 '13 at 19:45

This more a comment than an answer but will be too long for that. There are two basic approaches to the theory of finite measures---the topogical one and the one based on $\sigma$-algebras. At the core of the first approach lies the duality between $C^b(S)$ (the bounded, continuous functions on a completely regular space $S$) and the space $M^t(S)$ of Radon measures thereon, for the second that between $L^\infty(\Omega)$ (the bounded measurable functions on a measure space $(\Omega,\cal A)$ and the space $\cal M(\cal A)$ of finite $\sigma$-additive real valued-measures. All four of these spaces are Banach spaces in a natural way but these structures are not compatible with the above dualities. This suggests that one should try to provide the function spaces with suitable locally convex topologies which are compatible. In the case of $C^b(S)$ this means a suitable intrinsic locally convex topology which is complete and has the required dual space $M^t(S))$.
Due to the joint efforst of several mathematicians (in particular, R.C. Buck and the polish school, e.g., Orlicz and Wiweger) such a topology is known---it is the so-called strict topology (strictly speaking, this is not always complete, but it is so for most of the standard classes of topological spaces).

There is such a topology in the second case but I have been unable to find it in the literature and this is the reason that I am submitting this answer. It has various descriptions (in itself, in my opinion, a hint that it is a useful and natural topology). But for me the killer-diller fact is that it has the natural universal property for $\sigma$-additive measures with values in a Banach space (in contrast to the topological case, where we can embed $S$ into the space $M^t(S)$ with corresponding universal property), we can here embed the $\sigma$-algebra $\cal A$ into $L^\infty(\Omega)$ in the natural way.

We close with a few remarks.

$1$. We emphasise that it is important that the above topologies have intrinsic definitions (i.e., independent of the dual pairs). The fact that they then turn out to be topologies which can be defined via the duality---typically the Mackey topology---is then a theorem to be proved. This is, basically, the reason for
the relevance of all this to the OP which essentially asks for a characterisations of the weakly compact subsets of the dual of $L^\infty$.

$2$. Another reason for depositing this was to clarify the confusion about which is the relevant topology on the family of measures---it is clear from the formulation that the OP refers to the weak star topology of a subsset of the dual of $L^\infty$ with the topology referred to above.

$3$. If one tries to weaken the topology of a Banach space in a non-trivial way (as we are doing here) then there are contraints on the kind of space one can get. For example, one cannot get a barrelled space (closed graph theorem). This means that the resulting
topology
cannot be a member of one of the traditional classes of well-behaved LCS's (metrisable, inductive limits of Banach spaces, e.g.). Neither can they be nuclear (except, of course, in trivial cases). This, alas, seems to have been a hindrance to their acceptance into the main body of functional analysis. It is my belief that this is a great pity since they are precisely the tool required for extending and enriching the relationship between functional
analysis (duality theory) and measure theory (the Riesz representation theorem) which is one of the crown jewels of analysis.

What do you mean by universal property?Can you specify a bit more what you mean by compatibility in the first part?
–
andy teichMay 20 '13 at 9:31

Compatibility in the first part means that the corresonding dual spaces are too large. Thus the Banach space duals of the function spaces consist in both cases of the finitely additive measures, not the countably additive or Radon ones. The unversal property is that every countably additive meaure on the $\sigma$-algebra with values in a Banach space (for which see Diestel and Uhl "Vector measures") lifts to a unique continuous linear mapping on $L^\infty$ with the topology mentioned in my answer.
–
jbcMay 20 '13 at 10:06

By "compatibility" above I of course meant "non-compatibility". It is perhaps worth mentioning that in the topological situation, the universal property works in the other direction. $S$ embeds into $M^t(S)$ in such a way that every continuous, bounded function on $S$ with values in a Banach space lifts in a unique fashion to a continuous linear mapping with the appropirate (which, again, is not the norm).
–
jbcMay 20 '13 at 10:42